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Chemistry NEET Questions — Free MCQ Practice

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A buffer solution is made by mixing a weak acid HA (K_a = 10^–6) with its salt NaA in equal amounts. What should be the amount of acid or salt that should be added to make 90 ml of buffer solution of buffer capacity. 0.1 ?

A.

10 milli moles

B.

22 milli moles

C.

9 milli moles

D.

11 milli moles

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Explanation

For buffer capacity of 0.1 we should have $(\frac{c + 0 .1}{c - 0 .1})$ = 10

where is concentration of weak acid or salt in the buffer solution

So,c + 0.1 = 10 c – 1so9c = 1.1or c = $\frac{1 .1}{9}$

So, moles required for 90 solution = $\frac{1 .1}{9}$ × 90 × 10^–3 moles = 11 milli moles.

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A sample of water has a hardness expressed as 80 ppm of Ca²⁺. This sample is passed through an ion exchange column and the Ca²⁺ is replaced by H⁺. What is the pH of the water after it has been so treated? [Atomic mass of Ca = 40]

A.

3

B.

2.7

C.

5.4

D.

2.4

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Explanation

10⁶ ml water contains 80 gm of Ca²⁺ = $\frac{80}{40}$ moles = 2 moles of Ca²⁺ = 2 × 2 moles of H⁺ ions so 10³ ml of H₂O will have = 4 × 10^–3 moles of H⁺ ions

so pH = 3 – log 4 = 3 – 0.6 = 2.4.

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In the reaction COCl₂(_g) $\overset{}{⇌}$ CO(g) + CI₂(g) at 550°, when the initial pressure of CO & Cl₂ are 250 and 280 mm of Hg respectively. The equilibrium pressure is found to be 380 mm of Hg. Calculate the degree of dissociation of COC1₂ at 1 atm. What will be the extent of dissociation, when N₂ at a pressure of 0.4 atm is present and the total pressure is 1 atm.

A.

0.32 and no change

B.

0.32 and 0.4

C.

0.4 and 0.3

D.

In pressure of N₂ dissociation cannot takes place

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Explanation

CoCl₂ (g) $\overset{}{⇌}$ CO(g) + Cl₂ (g)

I.Pr – 250 280

Eq.pr x 250–x 280–x

x + 250 – x + 280 – x = 380

x = 150

K_p = 0.114

K_p = $\frac{P_{CO} \times P_{{Cl}_{2}}}{P_{{COCl}_{2}}}$

K_p = $\frac{α^{2} . p}{1 - α^{2}}$

0.114 = $\frac{α^{2} . 1}{1 - α^{2}}$

In presence of N₂ (constant pressure process)

K_p = $\frac{α^{2} \times 0 .6}{1 - α^{2}}$

0.114 = $\frac{α^{2} \times 0 .6}{1 - α^{2}}$

α = $\sqrt{\frac{0 .115}{0 .715}}$

α = 0.4

α–increases from 0.32 to 0.4.

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What is the value of pK_b (CH₃COOH^–) if $λ_{m}^{\infty}$ = 390 & $λ_{m}^{}$ = 7.8 for 0.04 of a CH₃COOH at 25°C

A.

9.3

B.

9.2

C.

4.7

D.

4.8

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Explanation

$α = λ m / λ m = 7.8 / 390 = 0.02 K_{a} = L α^{2} = 0.04 {(0.02)}^{2} = 1.6 \times 10^{- 5} 1 - α i s n e g l e c t e d b e c a u s e C H_{3} C O O H = w e a k e l e c t r o l y t e P k b = 14 - P k a = 14 - [- \log k a] P k b = 14 - [- \log (1.6 \times 10^{- 5})] P k b = 14 - 4.79 = 4.20$

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Assertion : It is difficult to distinguish the strengths of the strong acids such as

$HCl, H_{2} {SO}_{4}, {HNO}_{3}, HBr, Hl or {HClO}_{4}$ in dilute aqueous solutions.

Reason : In dilute aqueous solution all strong acids donate a proton to water and are essentially.

100% ionised to produce a solution containing $H_{3} O^{+}$ ions plus the anions of strong acid.

A. If both the assertion and the reason are true and the reason is a correct explanation of the assertion B. If both the assertion and reason are true but the reason is not a correct explanation of the assertion C. If the assertion is true but the reason is false D. If both the assertion and reason are false

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Explanation

(A) All are strong $(a = 1)$ (leveling effect).

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Assertion : 0.20 M solution of $NaCN$ is more basic than 0.20 M solution of $NaF$ .

Reason : 0.20 M solution of $NaCN$ is more basic than 0.20 M solution of ${CH}_{3} COONa .$

A. If both the assertion and the reason are true and the reason is a correct explanation of the assertion B. If both the assertion and reason are true but the reason is not a correct explanation of the assertion C. If the assertion is true but the reason is false D. If both the assertion and reason are false

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Explanation

(B) ${CN}^{-}$ is more basic than $F^{-} K_{a} (HCN) < K_{a} (HF)$

${CN}^{-}$ is more basic than ${CH}_{3} {COO}^{-} K_{a} (HCN) < K_{a} ({CH}_{3} COOH)$

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Assertion: A substance that can either act as an acid or a base is called ampholyte.

Reason: Bisulphide ion $({HS}^{-})$ and bicarbonate ion $({HCO}_{3}^{-})$ are ampholytes.

A. If both the assertion and the reason are true and the reason is a correct explanation of the assertion B. If both the assertion and reason are true but the reason is not a correct explanation of the assertion C. If the assertion is true but the reason is false D. If both the assertion and reason are false

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Explanation

(B) Ampholyte ion can undergo hydrolysis as well as ionization.

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The most suitable method of separation of 1:1 mixture of ortho and para-nitrophenols is

A.

sublimation

B.

chromatography

C.

crystallisation

D.

steam distillation

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Explanation

(d) Steam distillation is used to purify the substances which

(i) are volatile in steam but are immiscible with water.

(ii) possess sufficiently high vapour pressure at the boiling point of water.

(iii) contain non- volatile impurities.

The process of steam distillation can also be used to separate a mixture of two organic compounds one of which is steam volatile while the other is not. In ortho and para-nitrophenols, the latter is non-volatile, hence they are separated by steam distillation.

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Which of the following statement is true if the reaction quotient, Q is equal to 1 ?

A.

$∆ G = 0$

B.

$∆ G^{\circ} = 0$

C.

$∆ S^{\circ} = 0$

D.

$∆ G = ∆ G^{\circ}$

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Explanation

$∆ G = ∆ G^{\circ} + R T \ln Q$

If reaction quotient = 1

$∴ ∆ G = ∆ G^{\circ}$

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5mL of NHCl, 20 mL of N/2 H₂SO₄ and 30mL of N/3 HNO₃ are mixed together and volume made one litre. The normality of the resulting solution is:

A.

N/5

B.

N/10

C.

N/20

D.

N/40

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Explanation

(d) M eq. of HCl = 5 x1 =5;

M eq. of H₂SO₄ = 20 x (1/2) = 10;

M eq.of HNO₃ = 30 x (1/3) =10;

Thus, total M eq. of acid = 5+10+10=25

Total volume = 1000mL

Also M eq. = N x V,

N = 25/1000=1/40

Chemistry MCQs for NEET — Practice Questions with Answers

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